Point $M$ is the midpoint of side $BC$ of an acute—angled triangle $ABC$. The point $U$ is symmetric to the orthocenter $ABC$ relative to its circumcenter. The point $S$ inside triangle $ABC$ is such that $US = UM$. Prove that $SA + SB + SC + AM < AB + BC + CA$.

Express $\int_0^2 f(x)dx$ for any quadratic functions $f(x)$ in terms of $f(0),\ f(1)$ and $f(2).$

Given are $a_0,a_1, ... , a_n$, satisfying $a_0=a_n = 0$, and $a_{k-1} - 2a_k+a_{k+1}\ge 0$ for $k=0, 1, ... , n-1$. Prove that all the numbers are negative or zero.

Simultaneously from the same point of a circular route and in the same direction for two hours two bodies move evenly. The first body performs a complete rotation three minutes faster than the second body and exceeds it every $9$ minutes and $20$ seconds. Whenever the first body will overtake the other the second exactly at the starting point?

We say a nondegenerate triangle whose angles have measures $\theta_1$, $\theta_2$, $\theta_3$ is [i]quirky[/i] if there exists integers $r_1,r_2,r_3$, not all zero, such that \[r_1\theta_1+r_2\theta_2+r_3\theta_3=0.\] Find all integers $n\ge 3$ for which a triangle with side lengths $n-1,n,n+1$ is quirky. [i]Evan Chen and Danielle Wang[/i]

Six musicians gathered at a chamber music festival . At each scheduled concert some of these musicians played while the others listened as members of the audience . What is the least number of such concerts which would need to be scheduled in order to enable each musician to listen , as a member of the audience, to all the other musicians? (Canadian origin)

Determine all positive integers $N$ such that $$2^N-2N$$ is a perfect square. (Walther Janous)

In a game of ping-pong, the score is $4-10$. Six points later, the score is $10-10$. You remark that it was impressive that I won the previous $6$ points in a row, but I remark back that you have won $n$ points in a row. What the largest value of $n$ such that this statement is true regardless of the order in which the points were distributed?

$BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic.

Given is an acute triangle $ABC$ with $BC < CA < AB$. Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$. Perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at $M$. Prove that $CK + KM = BL + LM$.

Let $a,b$ be positive integers . How many integers numbers can be written in the form $ap+bq$, where $p,q$ are nonnegative integers and $p+q\le{2005}$ ?

Solve equation $$x^2-\sqrt{a-x}=a$$ where $x$ is real number and $a$ is real parameter

A regular tetrahedron $SABC$ of volume $V$ is given. The midpoints $D$ and $E$ are taken on $SA$ and $SB$ respectively and the point $F$ is taken on the edge $SC$ such that $SF: FC = 1: 3$. Find the volume of the pentahedron $FDEABC$.

Does there exist a nonzero algebraic number $\alpha$ with $|\alpha| \neq 1$ such that there exists infinitely many positive integers $n$ for which there's $\beta_n \in \mathbb{C}$ with $\beta_n \in \mathbb{Q}(\alpha)$ and $\beta_n^n = \alpha$?

Given is natural number $n$. Sasha claims that for any $n$ rays in space, no two of which have a common point, he will be able to mark on these rays $k$ points lying on one sphere. What is the largest $k$ for which his statement is true?

Show that if the sides $a, b, c$ of a triangle satisfy the equation \[2(ab^2 + bc^2 + ca^2) = a^2b + b^2c + c^2a + 3abc,\] then the triangle is equilateral. Show also that the equation can be satisfied by positive real numbers that are not the sides of a triangle.

Call a rational number $r$ [i]powerful[/i] if $r$ can be expressed in the form $\dfrac{p^k}{q}$ for some relatively prime positive integers $p, q$ and some integer $k >1$. Let $a, b, c$ be positive rational numbers such that $abc = 1$. Suppose there exist positive integers $x, y, z$ such that $a^x + b^y + c^z$ is an integer. Prove that $a, b, c$ are all [i]powerful[/i]. [i]Jeck Lim, Singapore[/i]

Let $P(x)$ be a nonzero polynomial with integer coefficients. Let $a_{0}=0$ and for $i \ge 0$ define $a_{i+1}=P(a_{i})$. Show that $\gcd ( a_{m}, a_{n})=a_{ \gcd (m, n)}$ for all $m, n \in \mathbb{N}$.

Inside the parallelogram $ABCD$, point $M$ is chosen, and inside the triangle $AMD$, point $N$ is chosen in such a way that $$\angle MNA + \angle MCB =\angle MND + \angle MBC = 180^o.$$ Prove that lines $MN$ and $AB$ are parallel.

Prove that in any set of $2000$ distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d $, such that \[ \left| \frac{a-b}{c-d} - 1 \right|< \frac{1}{100000}. \]

Find all pairs $(a,b,c)$ of positive integers such that $$\frac{a}{2^a}=\frac{b}{2^b}+\frac{c}{2^c}$$

Positive numbers $a_1,a_2,...,a_n, b_1, b_2,...,b_n$ satisfy the condition $a_1+a_2+...+a_n=b_1+ b_2+...+b_n=1$. Find the smallest possible value of the sum $$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+...+\frac{a_n^2}{a_n+b_n}$$ (V.Kolbun)

Let $A(x)=\lfloor\frac{x^2-20x+16}{4}\rfloor$, $B(x)=\sin\left(e^{\cos\sqrt{x^2+2x+2}}\right)$, $C(x)=x^3-6x^2+5x+15$, $H(x)=x^4+2x^3+3x^2+4x+5$, $M(x)=\frac{x}{2}-2\lfloor\frac{x}{2}\rfloor+\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}+\ldots$, $N(x)=\textrm{the number of integers that divide }\left\lfloor x\right\rfloor$, $O(x)=|x|\log |x|\log\log |x|$, $T(x)=\sum_{n=1}^{\infty}\frac{n^x}{\left(n!\right)^3}$, and $Z(x)=\frac{x^{21}}{2016+20x^{16}+16x^{20}}$ for any real number $x$ such that the functions are defined. Determine $$C(C(A(M(A(T(H(B(O(N(A(N(Z(A(2016)))))))))))))).$$ [i]2016 CCA Math Bonanza Lightning #5.3[/i]

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

Let $O$ be an interior point of an acute triangle $ABC$. The circles with centers the midpoints of its sides and passing through $O$ mutually intersect the second time at the points $K$, $L$ and $M$ different from $O$. Prove that $O$ is the incenter of the triangle $KLM$ if and only if $O$ is the circumcenter of the triangle $ABC$.